MAE 140 – Linear Circuits – Fall 2007
Midterm – Solution
Question 1 [Thevenin equivalent]
Regarding the following circuit:
3A
6Ω
4Ω
30 V +
−
A
8A
B
A and B using source
a) [4 marks] Determine the Thevenin equivalent as seen from terminals transformations only.
Transform the 3 A current source in parallel with the 4 Ω resistor.
30 V +
−
4Ω
12 V
+
−
6Ω
A
8A
B
Transform the 30 V voltage source in series with the 6 Ω resistor.
6Ω
+
−
4Ω
12 V
A
5A
8A
B
Combine the 5 A and the 8 A sources of current into a single 3 A source of current.
1
12 V
6Ω
+
−
4Ω
A
3A
B
Transform the 3 A current source in parallel with the 6 Ω resistor.
4Ω
12 V
+
−
6Ω
A
18 V
+
−
B
Combine the 12 V and the 18 V sources of voltage into a single 6 V source of voltage and associate
the resistor to obtain the Thevenin equivalent circuit.
10 Ω
A
6V
+
−
B
REMARK: You get 1 mark per correct transformation up to a max of 4 marks.
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b) [2 marks] How much power would be absorbed by a 10 Ω resistor connected between terminals
A and ?
B
Plug the resistor in the Thevenin equivalent.
10 Ω
A
10 Ω
6V
+
vR
iR
+
−
−
B
Use the voltage divider formula to conclude that
vR =
10
(−6) = −3 V,
10 + 10
iR =
so that the power absorbed by the resistor is pR = vR iR = 9/10 W.
3
vR
3
=− A
10
10
Question 2 [Superposition / Mesh analysis]
Regarding the following circuit:
+ v0 −
6Ω
30 V
6Ω
6Ω
i1
6Ω
+
−
i2
15 A
i3
a) [4 marks] Find v0 using superposition.
We will use superposition to compute
v0 = v1 + v2
where v1 is the solution when the current source is zero and v2 is he solution when the voltage
source is zero.
First case: Zero the current source, that is, replace the current source by an open circuit, and
redraw the circuit as follows.
A
6Ω
+
v1
6Ω
−
30 V
6Ω
+
−
B
6Ω
Voltage v1 can now be found using the voltage divider formula
v1 =
6
1
(vA − vB ) = (vA − vB )
6+6
2
4
and
vA − vB =
6//(6 + 6)
4
30 =
30 = 12 V
6 + [6//(6 + 6)]
6+4
where we used the fact that 6//(6 + 6) = 6 × 12/(6 + 12) = 4. That is
1
v1 = (vA − vB ) = 6 V.
2
Second case: Zero the voltage source, that is, replace the voltage source by a short circuit, and
redraw the circuit as follows.
i2
+
6Ω
v2
−
6Ω
6Ω
6Ω
15 A
Voltage v2 is
v2 = 6 i2
where can now be found using the current divider formula. Indeed
i2 =
1
6
1
1
+
6 6 + (6//6)
(−15) A = −
1
6
3
15 A = − 15 A = −9 A
1 1
5
+
6 9
where we used the fact that 6 + (6//6) = 6 + 3 = 9. That is
v2 = 6 i2 = −54V.
The final result is then
v0 = v1 + v2 = 6 − 54 = −48 V.
REMARK: Two marks per case: one for correct setup, another for correct answer.
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b) [4 marks] Formulate mesh-current equations for the circuit. Clearly indicate the equations to
be solved and the unknowns. Use the mesh currents indicated in the drawing.
Because the 15 A source of current is on a single mesh we have
i3 = −15 A.
(1)
Writting the mesh-current equations by inspection for meshes 1 and 2 we have
 
i1
6 + 6 + 6 −6 −6  
0
i2 =
−6
6 + 6 −6
30
i3
Using (1) we obtain the final form of the mesh-current equations
−90
18 −6 i1
,
i3 = −15 A.
=
−60
−6 12
i2
REMARK: You get 1 mark if you stated eq. (1) correctly. 1 mark if you stated mesh-current
equation for mesh 1 correctly. One more if mesh 2 is correct. One more mark if indicated the final
equations to be solved, either by substituting i3 or clearly indicating that meshes 1 and 2 must be
solved in the presence of this constraint.
c) [2 marks] Use the mesh-current equations to find v0 .
First note that
v0 = 6 i1 .
(2)
Then determine i1 by solving
1 2 1 −90
1 3 −1 −1 −90
2 1 −3
−8
i1
=
=
=
.
=
−60
1 3 −2
−9
i2
6 −1 2
30 1 3 −60
Hence
v0 = 6 i1 = −48 V.
REMARK: You get 1 mark if you stated eq. (2) correctly. 1 more mark if you solved for i1
correctly.
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Question 3 [OpAmp Circuit Analysis]
Regarding the following resistive OpAmp circuit:
is
vs
A
+
−
RN
+
−
R
B
R
a) [4 marks] Show that the current is = −vs /RN .
Label voltages and currents in the circuit
is
vs
RN
A vP
iP
iN
+
−
vN
+
v0
−
R
B
R
Because iN = 0, the resistors R form a voltage divider
vN =
1
R
v0 = v0 .
R+R
2
A implies
KCL at node is = iP +
(vP − v0 )
RN
These two equations combined with vP = vN = vs and iP = 0. yield
is =
(vP − v0 ) (vs − 2vs )
vs
=
=−
RN
RN
RN
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as desired.
A
REMARK: You get one mark for stating vN correctly. One more for stating KCL @ node .
One for using vP = vN = vs , i p = in = 0. One for getting the right answer.
A and B in the direction indib) [2 marks] Draw the equivalent circuit as seen from terminals cated by the arrow. Explain why this circuit is called a “negative resistance”?
Hint: Recall that an equivalent circuit is one which has the same v-i characteristic.
From part a)
vs = Rs is ,
Rs = −RN .
The above is Ohm’s Law for a “negative resistance” RN . Hence the name of the circuit. The
equivalent is simply
A
−RN
B
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Question 4 [OpAmp Circuit Analysis (bonus)]
[2 marks] Consider the connection of the “negative resistance” circuit of Question 3 in parallel
C
with a load RL as seen in the next circuit. What is the equivalent circuit as seen from terminals D Draw the circuit. What happens when RL = RN ?
and ?
Hint: Make use of Question 3.
RN
C
+
RL
−
R
D
R
Use the equivalent obtained in Question 3
C
−RN
RL
D
from where an equivalent can be obtained after associating RL and −RN in parallel
REQ = RL // − RN =
1
1
1
−
RL RN
=
RL RN
RN − RL
When RN → RL then REQ → ∞, that is, the equivalent becomes an open circuit.
REMARK: One mark for REQ and one for the limit case.
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